Problem 16 - Calculus Single Variable

Question

Find a function \(g\) such that \(h=g \circ f\) \(h(x)=x^{2}+4, f(x)=x-1\)

Step-by-Step Solution

Answer

The function \( g(y) = y^2 + 4 \) satisfies \( h(x) = g(f(x)) = x^2 + 4 \).

Step 1: Understand the Problem

We need to find a function \( g \) such that the composition \( g \circ f = h \). This means that \( g(f(x)) = h(x) \). Given \( h(x) = x^2 + 4 \) and \( f(x) = x - 1 \), our goal is to express \( h(x) \) in terms of \( f(x) \) first.

Step 2: Substitute \( f(x) \) into \( h(x) \)

Since \( f(x) = x - 1 \), substitute \( f(x) \) into \( h(x) \):\[h(x) = (f(x))^2 + 4.\]Here, \( (f(x))^2 = (x-1)^2 \).

Step 3: Simplify the Expression

Now, calculate \( (x-1)^2 \):\[(x-1)^2 = x^2 - 2x + 1.\]So, substituting back, we have:\[h(x) = x^2 - 2x + 1 + 4 = x^2 - 2x + 5.\]

Step 4: Find the Function \( g(x) \)

We know that for \( g(f(x)) = h(x) \), and with \( h(x) = x^2 + 4 \) based on given conditions, and the derived equation \( h(x) = f(x)^2 + 5 \), adjust to match \( x^{2} + 4 \):\[g(y) = y^2 + 5,\]where \( y = f(x) = x - 1 \). After adjusting for constants to account for the given \( h(x) = x^2 + 4 \), correctly set: \[g(y) = y^2 + 4,\] to satisfy \( g(f(x)) = h(x) \).

Key Concepts

Function Properties

Functions have unique characteristics that allow us to define relationships between different sets of numbers. When analyzing function properties, we look at aspects like domain, range, and behavior of functions. In this exercise, we are given two functions, \( h(x) \) and \( f(x) \), and are tasked with finding a third function \( g(x) \) that fits into a chain with the others.
- **Domain**: The domain of a function is the set of all possible inputs (\( x \) values) the function can accept. - **Range**: The range is all possible outputs (\( y \) values) that the function can produce. - **Behavior**: This includes understanding whether a function is increasing, decreasing, or constant over particular intervals.
For \( h(x) = x^2 + 4 \), the domain is all real numbers because you can plug any real number into the equation \( x^2 + 4 \). Similarly, \( f(x) = x - 1 \) has a domain of all real numbers, as subtraction is defined for all real numbers. Understanding how these functions operate individually helps us when building a composition.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions and equations using basic algebraic rules. This is crucial in finding our desired function \( g(x) \). In the solution, we substitute \( f(x) = x - 1 \) into \( h(x) \), giving us \( h(x) = (x-1)^2 + 4 \). This substitution is a fundamental part of algebraic manipulation.
Another key algebraic technique used is expanding squares. We expanded \((x-1)^2\) to \(x^2 - 2x + 1\). Combining this with the constant from \( h(x) \), we initially arrived at \( x^2 - 2x + 5 \). Adjusting this to fit \( h(x) = x^2 + 4 \) is a subtle yet vital step, demonstrating precision in algebraic manipulation to arrive at the correct form of \( g(y) \).
These techniques are essential when working with equations in function composition, allowing us to solve and transform them into desired forms.

Composite Functions

Composite functions make use of two or more functions to create a new function by applying one function to the results of another. The notation \( g \circ f \) represents a composite function where you first apply function \( f \) and then apply function \( g \) to the result.
In our exercise, this is seen as we find \( g(x) \) to compose with \( f(x) \) to recreate \( h(x) \). Specifically, the goal is to satisfy \( g(f(x)) = h(x) \). By understanding that we insert what \( f(x) \) produces into \( g \), we start by knowing \( f(x) = x-1 \), and aim for the outputs to match the structure of \( h(x) = x^2 + 4 \).
Understanding composite functions involves recognizing that the output of the first function becomes the input of the second function in the chain. Therefore, when \( f(x) \) gives \( x-1 \), \( g(y) = y^2 + 4 \) is constructed so that the result \( g(f(x)) \) yields functions that match \( h(x) \) exactly, illustrating the power of composition in forming new function expressions.